a, $/2x - 5/=4 $
\(⇔\left[ \begin{array}{l}2x - 5=4 \\2x-5=-4\end{array} \right.\)
\(⇔\left[ \begin{array}{l}2x=9\\2x=1\end{array} \right.\)
\(⇔\left[ \begin{array}{l}x=\dfrac{9}{2} \\x=\dfrac{1}{2} \end{array} \right.\)
Vậy $x=\dfrac{9}{2}$ $hoặc$ $x=\dfrac{1}{2}$
b, $\dfrac{1}{3}-$ $/\dfrac{5}{4}-2x/=$ $\dfrac{1}{4}$
$⇔/\dfrac{5}{4}-2x/=$ $\dfrac{1}{12}$
\(⇔\left[ \begin{array}{l}\dfrac{5}{4}-2x=\dfrac{1}{12} \\\dfrac{5}{4} -2x=\dfrac{-1}{12} \end{array} \right.\)
\(⇔\left[ \begin{array}{l} x=\dfrac{7}{12} \\x=\dfrac{2}{3}\end{array} \right.\)
Vậy $x=\dfrac{7}{12}$ hoặc $x=\dfrac{2}{3}$
c, $\dfrac{3}{4}-/2x+1/=$ $\dfrac{7}{8}$
$⇔/2x+1/=\dfrac{-1}{8}$
Vì $/2x+1/$ $\geq0$
⇒ Phương trình vô nghiệm.
d, $/x - 1,5/= 2$
\(⇔\left[ \begin{array}{l}x= \dfrac{7}{2} \\x=\dfrac{-1}{2} \end{array} \right.\)
Vậy $x=\dfrac{7}{2}$ hoặc $x=\dfrac{-1}{2}$
e, $⇒/x+1/=0$ và $y-3=0$
$⇒x=-1$ và $y=3$
